'||=' Ruby 中的运算符 [英] '||=' operator in Ruby

问题描述

有人可以向我解释以下 Ruby 代码的含义吗?(我在一个人的项目中看到了这个代码片段):

Could some one explain to me the meaning of the following Ruby code? (I saw this code snippet in one guy's project):

car ||= (method_1 || method_2 || method_3 || method_4)

上面的代码和下面的代码有什么区别?

What is the difference between the above code and the following code?

car = method_1 || method_2 || method_3 || method_4

---------更新--------------

好的，在阅读@Dave 的解释后，我明白了 ||= 运算符的含义，我的下一个问题是如果 method_2、method_3 和 method_4 返回一个值，哪个值会分配给car?(我想 car 最初是 nil)

Ok, I got the meaning of ||= operator after read @Dave's explanation, my next question is if both method_2, method_3 and method_4 return a value, which one's value will be assigned to car? (I suppose car is nil initially)

推荐答案

这是'条件赋值'的赋值运算符

It's an assignment operator for 'Conditional Assignment'

见这里 -> http://en.wikibooks.org/wiki/Ruby_Programming/语法/运算符

条件分配:

 x = find_something() #=>nil
 x ||= "default"      #=>"default" : value of x will be replaced with "default", but only if x is nil or false
 x ||= "other"        #=>"default" : value of x is not replaced if it already is other than nil or false

运算符 ||= 是表达式的简写形式:

Operator ||= is a shorthand form of the expression:

x = x || "default" 

看到OP的编辑后，这个例子只是这个的扩展，意思是:

After seeing OP's edit, the example is just an extension of this, meaning:

car = method_1 || method_2 || method_3 || method_4

会将method_1、method_2、method_3、method_4(按此顺序)的第一个非零或非假返回值分配给car，否则将保留其旧值.

Will assign the first non-nil or non-false return value of method_1, method_2, method_3, method_4 (in that order) to car or it'll retain its old value.

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